# Algebra 1

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solve: log8(w-6)=2-log8(w+15)
note: the 8 is to be little at the bottom right corner of log

• Algebra 1 -

since 64 = 8^2, then using base-8 logs, we have

log(w-6) = log(64)-log(w+15)
(w-6) = 64/(w+15)
(w-6)(w+15) = 64
w^2 + 9w - 154 = 0
w = 1/2 (-9±√697)
scrap the negative value (why?)

• Algebra 1 -

because the answer has got to be a real number which is not a negative number

• Algebra 1 -

how do you get 1/2 as a solution?

• Algebra 1 -

So technically there would be no solution right?

• Algebra 1 -

So, technically, the solution is

x = 1/2 (-9+√697) = 8.70

where does the 1/2 come from? Think back ... back ... back to Algebra I and the quadratic formula.

• Algebra 1 -

• Algebra 1 -

SO HOW DO YOU GET 1/2 FORM THAT?

• Algebra 1 -

Britt, when Steve and I went to school, the quadratic formula used to be
(-b ± √(b^2 -4ac)/(2a) or
(1/2) (-b ± √(b^2 -4ac)/a , thus the 1/2

I was not aware of any recent changes to the formula.

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